# Space elevator orbit below GSO but slow enough to be caught by an airplane

I've read a space elevator at GSO 35,800 km (actually above GSO to create the needed counterweight) is not feasible due to lack of current available strength of materials. Also that a tether at low Earth orbit 200 km is not feasible due to extremely high speeds in the atmosphere leading to very high drag and impossible speeds at which to 'catch' the tether for delivery of payloads/people.

Is there an orbit in between these two extremes that would allow a tether to be short enough to be feasible from a strength and from an atmospheric speed/drag standpoint and what is that orbit?

• GSO isn't enough for a space elevator itself: there needs to be a counterbalance above GSO for the concept to work. – Baldrickk Aug 1 '17 at 9:53

Firstly, while the idea sounds good, it turns out it can't work at all even if we did have the materials for the tether and a plane. The reason is that a space elevator doesn't just have one counterweight! It has two. The first one is the obvious one, out beyond GSO. The second one is the earth! In order for our space elevator to be effective with a plane, the plane is going to have to weigh a very large amount if we want to send reasonable amounts of mass to the top. But that is the boring answer, with lots of ifs and buts. Anything is possible if you drop enough constraints. But say we did have a big enough plane, and could build your space elevator.

Lets do the math to find out what altitude this orbit could be at!

Assume the airplane travels over the surface at 240 m/s along the equator. Also at the equator, the earth rotates at $2\pi r_{earth}/24\mathrm{hr}=460\mathrm{m/s}$. So now our airplane is travelling around the earth at 700m/s in an earth-centered-inertial frame. Our airplane pretty close to the surface of the earth, lets say 10km altitude. This gives a radius from the center of 6,381km.

Now for the orbit. First, we calculate the angular velocity: $$\omega=\frac{v}{2\pi r_{plane}}=1.097\cdot10^{-4}\mathrm{rad/sec}$$ And finally we use the formula for geostationary orbit: $$r_{GSO}=\sqrt[3]{\frac{GM}{\omega^2}}-r_{earth}=25744\mathrm{km}$$

Sadly, this is only 30% less than the original value, and still vastly out to space compared to the ISS.

To bring the orbit down to say, 3629km (still much further than the ISS!), the plane would have to travel at $$\sqrt{\frac{GM}{r^3}}\cdot r_{plane}-v_{earth}=3570\mathrm{m/s}$$

Ouch. And like Baldrickk said, the counterweight has to be much further out than station at the top of the space elevator.

• I've never heard of the bottom counter weight, usually just that you need a counter weight at the top of the orbit and it tapers down as you approach the surface. And that you need to add more strength as you add 'any' weight to the elevator in the form of payload, and for the actual motion up the elevator as well. Awesome, I'm actually surprised it cut that much off the top! I thought I might gain 10% not 30%, but I'm sure a 25,744 km elevator is still beyond our current abilities. (I was hoping for 15 km from the ground, but 5 from 25k is negligible really!) – Brooks Nelson Aug 1 '17 at 16:15
• Indeed, we don't think about the bottom counterweight really. It is only there for when there is no elevator weighing the system down. But the reality is that there needs to be tension no matter what with this system, and the tension will be modest to large at the bottom and astronomically high at the orbiting counterweight, hence the technical difficulties. – Quietghost Aug 1 '17 at 16:22

It can be done if you back up and change some of your ideas about what you're building. Forget the space elevator per se, Quietghost demolished that. Space elevators must have a net upwards pull or they fall down and that upward pull is only generated by mass beyond synchronous orbit.

However, there is also the concept called the rotovator. Take a cable and put it in orbit. It is spun such that the top of the cable is moving forward and the bottom is moving backwards relative to it's orbit. If you build it big enough you can make the bottom part move backwards as fast as the center of mass is going forward--thus to an observer on Earth it would appear to slow as it dipped down, come to a momentary rest and then climb back up and speed up.

Now, if you have god-like ability to make things precise it could come down and pick up a payload that's sitting on surface. Since getting it ever so slightly wrong would be catastrophic, though, in practice you are going to intercept the cable somewhere in flight.

Note that this system is not tethered to produce a net outward force. This means there's no free lunch in lifting payloads into space, for every Newton you use to speed up your payload you must also exert a Newton slowing something down. You also must manage the rotational speed of the rotovator.

A lifter that must bring down as much as it brings up at first glance might seem to not have much value but that's not the case--what comes down need not be what goes up. Put a mining base on the moon, pack up your refinery slag and use that as the mass to be brought back down.

• Refueling rocket engines of the rotovator that would keep it spinning and flying would be vastly cheaper, as fuel can be delivered by the same means as payload - by airplane. – SF. Aug 1 '17 at 22:59
• @SF Not if you're using chemical rockets, the exhaust velocity is nowhere near high enough. – Loren Pechtel Aug 1 '17 at 23:20
• I'm happy that you mention catching payload from higher orbits can restore a rotovator's momentum. But instead of sending slag for your up momentum, I would send hydrogen/oxygen bipropellent (If there are indeed rich ice deposits in the cold traps). – HopDavid Aug 2 '17 at 4:26
• @HopDavid Why?!?! You get more energy out of mass by discarding it at the low point of the rotovator than by using it as rocket fuel at the center. Use the hydrogen/oxygen to go somewhere you can't get with the rotovator. – Loren Pechtel Aug 2 '17 at 4:56
• "Use the hydrogen/oxygen to go somewhere you can't get with the rotovator." Sending down lunar bipropellent would do both. Catching it would provide the LEO rotovator up momentum. After it's caught it could be stored in a LEO propellent depot. Payloads dropped from the tether could have some reaction mass to further slow their re-entry. – HopDavid Aug 2 '17 at 14:27